Question

1 kg of an aqueous solution of Sucrose is cooled and maintained at -4°C. How
much ice will be separated out if the initial molality of the solution is 0.75?
K (H2O) = 1.86 Kg mol-'K.​

Answer 1

Sucrose is a non-electrolyte, hence i = 1

Molecular weight = m = 342 gm/mol

Molality of solution = 0.75 m

= 0.75 mol/kg solvent

= 0.75×342 gm

= 256.5 gm sucrose

Weight of H2O(solvent) present in 1kg solution:

w = 1000 - 256.5 = 743.5 gm

Since, there is depression in freezing point,

∆T = k×i×w/m×1000/W

Where, W = weight of solvent

w = weight of solute

or, 4 = 1.86×1×204.1/342×1000/W

Therefore, W = 277.55 gm

That is, weight of solvent required to maintain this solution at -4℃ is W = 277.55 gm

Hence, rest weight of H2O will convert into ice.

Hence, amount of ice formed = 743.5 - 277.55 = 465.95 gm

Thus, 465.95gm of ice will be separated out.